Q20. - Algebra 2

Question

Solve each system of equations 

$\begin{array}{l}2r+s+t=7\\ r+2s+t=8\\ r+s+2t=11\end{array}$

Step-by-Step Solution

Answer

The solution set of the given system of equations is $\left(\frac{1}{2},\frac{3}{2},\frac{9}{2}\right)$.

Step 1 – Use the elimination method to get the system of equations in two variables.

Subtract the equation $r+2s+t=8$ from $2r+s+t=7$:

$\begin{array}{l}\underset{\_}{\begin{array}{l}2r+s+t=7\\ (-)r+2s+t=8\end{array}}\\ r-s+0=-1\end{array}$

So, the resultant equation is $r-s=-1$.

 Next, multiply the equation $r+2s+t=8$ by $2$and subtract the new resultant equation from $r+s+2t=11$

$\begin{array}{l}\underset{\_}{\begin{array}{l}r+2s+t=8\\ r+s+2t=11\end{array}}\begin{array}{c}\text{multiply by 2}\to \\ \end{array}\underset{\_}{\begin{array}{l}2r+4s+2t=16\\ (-)r+s+2t=11\end{array}}\\ r+3s+0=5\end{array}$

So, the resultant equation is $r+3s=5$.

.

Step 2 – Use the elimination method to solve the system of two equations.

Multiply $r-s=-1$by 3 and add the new resultant equation from $r+3s=5$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}r-s=-1\\ r+3s=5\end{array}}\begin{array}{c}\text{multiply by 3}\\ \end{array}\underset{\_}{\begin{array}{l}3r-3s=-3\\ r+3s=5\end{array}}\\ 4r+0=2\end{array}$

Solve $4r=2$ for r

$\begin{array}{l}4r=2\\ \frac{4r}{4}=\frac{2}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Divide both sides by 4}\\ r=\frac{1}{2}\end{array}$

Step 3 – Find the values of s and t .

Substitute $r=\backslash frac\left\{1\right\}\left\{2\right\} in r-s=-1$ and find the value of $s$.

$\begin{array}{l}r-s=-1\\ \frac{1}{2}-s=-1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute }\frac{1}{2}\text{ for }r\\ -s=-\frac{3}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract }\frac{1}{2}\text{ from both sides}\\ s=\frac{3}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Divide both sides by -1}\end{array}$

Substitute $r=\backslash frac\left\{1\right\}\left\{2\right\}$,$s=\backslash frac\left\{3\right\}\left\{2\right\} in 2r+s+t=7$ and find the value of $t$

$\begin{array}{l}2r+s+t=7\\ 2\left(\frac{1}{2}\right)+\frac{3}{2}+t=7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute }\frac{1}{2}\text{ for }r,\frac{3}{2}\text{ for }s\\ 1+\frac{3}{2}+t=7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ t+\frac{5}{2}=7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ t=\frac{9}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract }\frac{5}{2}\text{from both sides}\end{array}$

Hence, the solution of the given system of equations is$\left(r,s,t\right)=\left(\frac{1}{2},\frac{3}{2},\frac{9}{2}\right)$.

.